Deriving the Area of a Circular Sector Using Integration

[Noriele Cruz]

I am bothered on how the area of circular sector was derived using integration. I have a solution but I don't know how it will go and come up with the formula:

A = 1/2 sr and A = 1/2 r² θ

Can someone show me the solution for the derivation of the two formula?

 

[blue_leaf77]

A basic question, what do you know about an integral?

 

[SteamKing]

I am bothered on how the area of circular sector was derived using integration. I have a solution but I don't know how it will go and come up with the formula:

A = 1/2 sr and A = 1/2 r² θ

Can someone show me the solution for the derivation of the two formula?

It's an elementary proof if you use polar coordinates.

Examine the small sector in the figure below:

​

You can assume for a small angle dθ that the sector is roughly an isosceles triangle, where each side measures r units and the base is r ⋅ dθ units long.

The area of this triangle is dA ≈ (1/2) ⋅ r × r ⋅ dθ = r² dθ / 2, using the formula for the area of a triangle.

If you want to find the area A of a circular sector between the angles θ1 and θ2, then

$$A = \int_{θ1}^{θ2} \,dA = \int_{θ1}^{θ2} \frac{1}{2} r^2\,dθ = \frac{1}{2} r^2 ⋅ (θ2 - θ1)$$

r ⋅ (θ2 - θ1) is also equivalent to the arc length s, and $A = \frac{1}{2} r ⋅ s$

Q.E.D.

 

[Noriele Cruz]

I have done it by means of basic integration of plane areas. I divide the figure into two parts as A₁ and A₂ for computing the area.

It is somehow like this (I can't find a picture of my solution). I draw a triangle from A perpendicular to CB. Now, the sector is divided into a triangle and segment. The triangle represents A₁ and segment represents A₂. Can you show me your solution in my way of derivation?

 

[SteamKing]

I have done it by means of basic integration of plane areas. I divide the figure into two parts as A₁ and A₂ for computing the area.

It is somehow like this (I can't find a picture of my solution). I draw a triangle from A perpendicular to CB. Now, the sector is divided into a triangle and segment. The triangle represents A₁ and segment represents A₂. Can you show me your solution in my way of derivation?

No, because that's a really difficult way to do this problem.

There's nothing wrong with switching to integration using polar coordinates when that method makes the calculation much easier.

 

[Noriele Cruz]

I am finish in integrating it with values.. I am stuck on how the integration will cancel out to come up with only 1/2 r² θ for the area.

What do you mean by polar coordinates? I think I work with cartesian coordinates in integrating?

 

[SteamKing]

I am finish in integrating it with values.. I am stuck on how the integration will cancel out to come up with only 1/2 r² θ for the area.

r is a constant with respect to theta. What happens when you integrate constants?

What do you mean by polar coordinates? I think I work with cartesian coordinates in integrating?

I mean polar coordinates:

https://en.wikipedia.org/wiki/Polar_coordinate_system

If you haven't studied polar coordinates, there is a certain mathematical background which is lacking in your education. You should correct this if you are going to study calculus.

 

[Noriele Cruz]

I know how to integrate constants. I just don't know how to simplify my solution to 1/2 r² θ.

We had study about polar coordinates and I know it, I just don't know how to make equations for the integration process.

 

[SteamKing]

I know how to integrate constants. I just don't know how to simplify my solution to 1/2 r² θ.

You'll have to show your work where you got stuck.

We had study about polar coordinates and I know it, I just don't know how to make equations for the integration process.

I show you how and explained it in Post #3.

 

[Noriele Cruz]

I got this part: πr² / 4 - r²/2 arcsin (cos θ)

Can you show me how, please?

 

[Noriele Cruz]

This is my solution:

 

[SteamKing]

I got this part: πr² / 4 - r²/2 arcsin (cos θ)

I'm not even sure what this expression is supposed to represent.

If it's supposed to be the area of a circular sector subtending an angle of 45°, it's way more complicated than it needs to be (and I haven't checked for any mistakes).

https://en.wikipedia.org/wiki/Circular_sector

 

[Noriele Cruz]

Have you checked my solution in the attached file?

For the formula A = 1/2 sr, how it was derived using integration? Sorry for so many questions, I just want to clear up my mind.

 

[SteamKing]

Have you checked my solution in the attached file?

I looked at it. It's way too complicated to spend time checking, especially when I have shown you in detail how to obtain a much simpler formula for the area of a circular sector, and showed you documentary proof of the validity of the simple formula.

For the formula A = 1/2 sr, how it was derived using integration? Sorry for so many questions, I just want to clear up my mind.

Again, please refer to Post #3. I showed how this formula is derived from the alternate formula for the area of a circular sector.

If you are not going to read the posts I put up, there's no point in continuing this thread.

 

[Noriele Cruz]

It's an elementary proof if you use polar coordinates.

Examine the small sector in the figure below:

​

You can assume for a small angle dθ that the sector is roughly an isosceles triangle, where each side measures r units and the base is r ⋅ dθ units long.

The area of this triangle is dA ≈ (1/2) ⋅ r × r ⋅ dθ = r² dθ / 2, using the formula for the area of a triangle.

If you want to find the area A of a circular sector between the angles θ1 and θ2, then

$$A = \int_{θ1}^{θ2} \,dA = \int_{θ1}^{θ2} \frac{1}{2} r^2\,dθ = \frac{1}{2} r^2 ⋅ (θ2 - θ1)$$

r ⋅ (θ2 - θ1) is also equivalent to the arc length s, and $A = \frac{1}{2} r ⋅ s$

Q.E.D.

How come my picture of my integration of sector can be drawn like this? Is this also the same with mine?

 

[SteamKing]

How come my picture of my integration of sector can be drawn like this? Is this also the same with mine?

All circular sectors look alike.

The dA in the figure is actually drawn for the case where r is not a constant but instead varies with the angle θ.

For a circular sector, r is constant, and integrating r and θ is not necessary to find the area of the sector.

 

[Noriele Cruz]

How about the curve in the part of the sector? If I'm not mistaken, the figure you posted is somehow a triangle rather than a sector with curve part.
Where is it?

 

[SteamKing]

How about the curve in the part of the sector? If I'm not mistaken, the figure you posted is somehow a triangle rather than a sector with curve part.
Where is it?

In the limit, as dθ → 0, then the arc portion of the sector is naturally going to appear more and more like a straight line segment, which is why the area of this tiny sector can be approximated using the formula for the area of a triangle.

To find the area of circular sector ABC, as shown in Post #4, that sector would be subdivided into many little slivers each sweeping out an angle dθ. The integration process adds up the areas dA of these segments as dθ → 0, producing a finite value A for the area of sector ABC.

 

[geoffrey159]

In general, the area of a sector bounded by a curve $C$ defined by polar radius $\rho(\theta)$, such that $C = \rho([\theta_1,\theta_2])$ , is $A= \frac{1}{2} \int_{\theta_1}^{\theta_2} \rho^2(\theta) \ d\theta$ . In your case, $\rho(\theta)$ is constant and equal to $r$ since your sector is circular.